Question
Two arithmetic progression have the same common difference. The difference between their $100^{\text {th }}$ terms is 100 , What is the difference between their $1000^{\text {th }}$ terms?
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Answer
Let the two A.P. is be $a _1, a _2, a _3, \ldots .$. and $b _1, b_2, b_3, \ldots .$.
$a_n=a_1+(n-1) d \text { and } b_n=b_1+(n-1) d$
Since common difference of two equations is same given difference between $100^{\text {th }}$ terms is $100$
$a_{100} - b_{100} = 100$
$a_1 + (99) d - b_1 - 99d = 100$
$a_1 - b_1 = 100 .....(i)$
Difference between. 1000th terms is
$a_{1000} - b_{1000} = a_1 + (1000 - 1)d - (b_1 + (1000 - 1)d)$
$= a_1 + 999d - b_1 - 999d$
$= a_1 - b_1$
$= 100$ (from (1))
$\therefore$ Hence difference between $1000^{\text {th }}$ terms of two $A.P$. is $100.$
$a_n=a_1+(n-1) d \text { and } b_n=b_1+(n-1) d$
Since common difference of two equations is same given difference between $100^{\text {th }}$ terms is $100$
$a_{100} - b_{100} = 100$
$a_1 + (99) d - b_1 - 99d = 100$
$a_1 - b_1 = 100 .....(i)$
Difference between. 1000th terms is
$a_{1000} - b_{1000} = a_1 + (1000 - 1)d - (b_1 + (1000 - 1)d)$
$= a_1 + 999d - b_1 - 999d$
$= a_1 - b_1$
$= 100$ (from (1))
$\therefore$ Hence difference between $1000^{\text {th }}$ terms of two $A.P$. is $100.$
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